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1.hdu3377:求最大流模板

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define qwq(x) for(edge *o=head[x];o;o=o->next)
int read(){
	int x=0;char c=getchar();bool f=true;
	while(!isdigit(c)) {
		if(c=='-') f=false;c=getchar();
	}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return f?x:-x;
}
const int nmax=105;
const int inf=0x7f7f7f7f;
struct edge{
	int to,cap;edge *next,*rev;
};
edge edges[10005],*head[nmax],*pt;
void add(int u,int v,int d){
	pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
	add(u,v,d);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];
}
int cnt[nmax],h[nmax];
edge *p[nmax],*cur[nmax];
int maxflow(int s,int t,int n){
	clr(cnt,0);cnt[0]=n;clr(h,0);
	int flow=0,a=inf,x=s;edge *e;
	while(h[s]<n){
		for(e=cur[x];e;e=e->next)
		  if(e->cap>0&&h[e->to]+1==h[x]) break;
		if(e){
			p[e->to]=cur[x]=e;a=min(a,e->cap);x=e->to;
			if(x==t){
				while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
				flow+=a;a=inf;
			}
		//	flow+=a;a=inf;
		}else{
			if(!--cnt[h[x]]) break;
			h[x]=n;
			for(e=head[x];e;e=e->next) 
			  if(e->cap>0&&h[x]>h[e->to]+1)
			     h[x]=h[e->to]+1,cur[x]=e;
			cnt[h[x]]++;
			if(x!=s) x=p[x]->rev->to;
		}
	}
	return flow;
}
int main(){
	int t=read(),n,m,u,v,d;
	rep(i,t){
		clr(head,0);pt=edges;
		n=read(),m=read();
		rep(j,m) u=read(),v=read(),d=read(),adde(u,v,d);
		/*rep(j,n){
			qwq(j) printf("%d %d  ",o->to,o->cap);
			printf("
");
		}*/
		printf("Case %d: %d
",i,maxflow(1,n,n));
	}
	return 0;
}

  poj1273:。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();bool f=true;
	while(!isdigit(c)) {
		if(c=='-') f=false;c=getchar();
	}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=205;
const int inf=0x7f7f7f7f;
struct edge{
	int to,cap;edge *next,*rev;
};
edge edges[405],*pt,*head[nmax],*cur[nmax],*p[nmax];
int cnt[nmax],h[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
	add(u,v,d);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];
}
int maxflow(int s,int t,int n){
	clr(cnt,0);cnt[0]=n;clr(h,0);
	int flow=0,a=inf,x=s;edge *e;
	while(h[s]<n){
		for(e=cur[x];e;e=e->next) if(e->cap>0&&h[e->to]+1==h[x]) break;
		if(e){
			a=min(a,e->cap);p[e->to]=cur[x]=e;x=e->to;
			if(x==t){
				while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
				flow+=a;a=inf;
			}
		}else{
			if(!--cnt[h[x]]) break;
			h[x]=n;
			for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1) h[x]=h[e->to]+1,cur[x]=e;
			cnt[h[x]]++;
			if(x!=s) x=p[x]->rev->to;
		}
	}
	return flow;
}
int main(){
	int n,m,u,v,d;
	while(scanf("%d%d",&m,&n)==2){
		clr(head,0);pt=edges;
		rep(i,m) u=read(),v=read(),d=read(),adde(u,v,d);
		printf("%d
",maxflow(1,n,n));
	}
	return 0;
}

  poj3281:建图即可。s->F->k->k'->D->t;

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();bool f=true;
	while(!isdigit(c)) {
		if(c=='-') f=false;c=getchar();
	}
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return f?x:-x;
}
const int nmax=405;
const int maxn=50000;
const int inf=0x7f7f7f7f;
struct edge{
	int to,cap;edge *next,*rev;
};
edge edges[maxn],*pt,*head[nmax],*cur[nmax],*p[nmax];
int cnt[nmax],h[nmax];
void add(int u,int v,int d){
	pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
	add(u,v,d);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];
}
int maxflow(int s,int t,int n){
	clr(cnt,0);cnt[0]=n;clr(h,0);
	int flow=0,a=inf,x=s;edge *e;
	while(h[s]<n){
		for(e=cur[x];e;e=e->next) if(e->cap>0&&h[e->to]+1==h[x]) break;
		if(e){
			a=min(a,e->cap);p[e->to]=cur[x]=e;x=e->to;
			if(x==t){
				while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
				flow+=a;a=inf;
			}
		}else{
			if(!--cnt[h[x]]) break;
			h[x]=n;
			for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1) h[x]=h[e->to]+1,cur[x]=e;
			cnt[h[x]]++;
			if(x!=s) x=p[x]->rev->to;
		}
	}
	return flow;
}
int main(){
	pt=edges; clr(head,0);
	int n=read(),F=read(),D=read(),tf,td,s=0,t=2*n+F+D+1,tmp;
	rep(i,n){
		tf=read(),td=read();
		rep(j,tf) tmp=read(),adde(tmp,F+i,1);;
		rep(j,td) tmp=read(),adde(F+n+i,F+n+n+tmp,1);
		adde(F+i, F+n+i,1);
	}
	rep(i,F) adde(s,i,1);
	rep(i,D) adde(F+n+n+i,t,1);
	printf("%d
",maxflow(s,t,t+1));
	return 0;
}

2.最优匹配可以用最小费用最大流做。。。最大匹配可以用最大流做。。。不知道时间复杂度比较如何。。。应该是匈牙利比较快吧。。。

3.

原文地址:https://www.cnblogs.com/fighting-to-the-end/p/5656157.html